If $$\sigma$$ is the standard deviation of $$x_1, x_2, ... ... x_n$$ then the standard deviation of $$9 + 3x_1, 9 + 3x_2, ..... ..... 9 + 3x_n$$ is

Let the mean of the first series be M
Hence, M = $$\frac{x_1+x_2+......+x_n}{n}$$

Now, for the second series, let its mean be M'
Hence, M' = $$\frac{3\cdot\left(3n+x_1+x_2+......+x_n\right)}{n}=9+3M$$

Now, respective deviation of the individual terms from the mean M' are:
3M - 3$$x_1$$ = 3(M-$$x_1$$)
3M - 3$$x_2$$ = 3(M-$$x_2$$)
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3M - 3$$x_n$$ = 3(M-$$x_n$$)

Variation = $$\frac{\left[9\cdot\left(M-x_1^{ }\right)\left(M-x_1\right)+9\cdot\left(M-x_2\right)\left(M-x_2\right)+....+9\left(M-x_n\right)\left(M-x_n\right)\right]}{n}$$
Standard deviation = Square root of this Variance = 3*$$\sigma\ $$